0.01 M solution of KCl and `CaCl_(2)` are separately prepared in water. The freezing point of KCl is found to be `-2^(@)C`. What is the freezing point of `CaCl_(2)` aq. Solution if it is completely ionized?

Answer

`-3^(@)C`

Explanation

i. for KCl= 2 , i.for `BaCl_(2)`
`Delta T_(f) (KCl) = 2 xx k_(f) xx 0.01`
`Delta T_(f) (BaCl_(2)) = 3 xx k_(f) xxx 0.01`
`(Delta T_(f)(KCl))/(Delta T_(f) (BaCl_(2)) = (3)/(2) xx Delta T_(f) (KCl) = (3)/(2) xx 2 = 3^(@)`
`therefore` Freezing point of `BaCl_(2) = - 3^(@)C`